Physics MCQ Quiz - Objective Question with Answer for Physics - Download Free PDF

Concept:

de Broglie Wavelength of an Electron in an Electric Field:

The de Broglie wavelength of a particle is given by:

λ = h / mv,

where:

h: Planck's constant

m: Mass of the particle

v: Velocity of the particle

When an electron of mass m enters a uniform electric field, it experiences a force, causing its velocity to change over time. This affects its de Broglie wavelength.

The velocity of the electron after time t in the electric field can be determined from the equation:

v = v₀ + (eE / m) t,

where:

e: Charge of the electron

E: Electric field strength

t: Time

Substituting the velocity into the de Broglie wavelength equation gives the expression for the wavelength after time t.

Calculation:

Given:

Initial de Broglie wavelength: λ₀ = h / m v₀

The velocity of the electron after time t is:

⇒ v = v₀ + (eE / m) t

Substituting this into the de Broglie wavelength equation:

⇒ λ' = h / m (v₀ + (eE / m) t)

⇒ λ' = λ₀ / √(1 + (eE t)² / m² v₀²)

∴ The value of the de Broglie wavelength after time t is: λ₀ / √(1 + (eE t)² / m² v₀²)

CONCEPT:

• Photoelectric effect: When the light of a sufficiently small wavelength is incident on the metal surface, electrons are ejected from the metal instantly. This phenomenon is called the photoelectric effect.

Photo-electric cell

• The photo-electric cell works on the principle of the photoelectric effect.
  • The photo-electric cell consists of an evacuated glass tube containing two electrodes emitter (C) and Collector (A).
  • The emitter is always kept at a negative potential.
  • The collector is made of a metal rod and is always kept at a positive potential.
  • When the light of frequency more than the threshold frequency of material of emitter is made incident on the emitter, photo-emission takes place.
  • Then the photo­electrons are attracted to the collector which is positive with respect to the emitter. Thus current flows in the circuit. If the intensity of incident radiation is increased the photoelectric current increases.

Calculation:

E = \(\frac{1240}{\lambda}=\frac{1240}{550} \simeq 2.25\)

Concept:

The de-Broglie wavelength of a particle is given by the formula:

\(λ = \frac{h}{mv}\)

Where:

• λ is the de-Broglie wavelength
• h is Planck's constant
• m is the mass of the electron
• v is the velocity of the electron

In the Bohr model of the hydrogen atom, the radius of the orbit is given by:

\(r_n = n^2 \times r_1\)

Where:

• \(r_n\) is the radius of the electron's orbit at the n -th level
• \(r_1\) is the radius of the first orbit (ground state, \(r_1 = 5.3 \times 10^{-11} m\))
• \(n\) is the principal quantum number

Calculation:

The de-Broglie wavelength of the electron is inversely proportional to the momentum, which depends on the radius of the orbit. Since the momentum is proportional to the radius (for a given energy level), the ratio of the de-Broglie wavelengths will be the inverse of the ratio of the radii:

\(\frac{λ_{\text{ground}}}{λ_{\text{excited}}} = \frac{r_{\text{excited}}}{r_{\text{ground}}}\)

Substitute the given values:

\(\frac{r_{\text{excited}}}{r_{\text{ground}}} = \frac{8.48 \times 10^{-10}}{5.3 \times 10^{-11}} = 16\)

Conclusion:

∴ The ratio of the de-Broglie wavelengths is 16.

Thus, the correct answer is: (4) 16

Concept:

Photoelectric Effect:

• In the photoelectric effect, electrons are ejected from the surface of a material when it is illuminated with light of frequency above a certain threshold.
• The energy of the incident photons is given by: E = hν, where h is Planck's constant and ν is the frequency of the light.
• The equation for the photoelectric effect can be written as: eVₒ = hν − φ, where:
  • e is the charge of the electron,
  • Vₒ is the stopping (cut-off) voltage,
  • h is Planck's constant,
  • ν is the frequency of incident light, and
  • φ is the work function of the material.
• When the cut-off voltage (Vₒ) is plotted against the frequency (ν), the slope of the graph is given by the relation: slope = h/e, where:
  • h is Planck's constant,
  • e is the charge of the electron (approximately 1.6 × 10⁻¹⁹ C).

Calculation:

Given,

Planck's constant, h = 6.63 × 10⁻³⁴ J·s

Charge of the electron, e = 1.6 × 10⁻¹⁹ C

Using the formula for the slope:

slope = h / e

slope = (6.63 × 10⁻³⁴) / (1.6 × 10⁻¹⁹)

slope = 4.14 × 10⁻¹⁵ V·s

∴ The slope of the graph is 4.14 × 10⁻¹⁵ V·s.
Hence, the correct option is 1) 4.14 × 10⁻¹⁵ V·s.

Concept:

The energy of a photon is given by E = hν = hc/λ. The maximum kinetic energy of the photoelectrons is given by the photoelectric equation:

K.E. = hf - φ

where φ is the work function of the metal.

Calculation:

Light of wavelength λA and λB falls on two identical metal plates A and B respectively. The maximum kinetic energy of photoelectrons is KA and KB respectively, with λA = 2λB.

For plate A, with wavelength λ_A:

⇒ K_A = hc/λ_A - φ

For plate B, with wavelength λ_B:

⇒ K_B = hc/λ_B - φ

Given λ_A = 2λ_B, substitute λ_A:

⇒ K_A = hc/(2λ_B) - φ

Since λ_A = 2λ_B, we have:

⇒ K_A = (1/2)hc/λ_B - φ

⇒ K_A = (1/2)(hc/λ_B - 2φ) + φ

From K_B equation:

⇒ K_B = hc/λ_B - φ

Since hc/λ_B - φ = K_B:

⇒ K_A = (1/2)(K_B + φ)

Since φ is the same for both metals:

⇒ K_A = (1/2)K_B + (1/2)φ

Therefore, the correct relation is:

∴ K_A = (1/2)K_B + (1/2)φ, indicating K_A is less than K_B.

CONCEPT:

• Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P = \frac{W}{t} = VI = {I^2}R = \frac{{{V^2}}}{R}\)

Where V = Potential difference, R = Resistance and I = current.

CALCULATION:

Given - Potential difference (V) = 220 V, power of the bulb (P) = 100 W and actual voltage (V') = 110 V

• The resistance of the bulb can be calculated as,

​\(\Rightarrow R=\frac{V^2}{P}=\frac{(220)^2}{100}=484 \,\Omega\)

• The power consumed by the bulb.

\(\Rightarrow P=\frac{V^2}{R}=\frac{(110)^2}{484}=25 \,W\)

CONCEPT:

Galvanometer:

• A galvanometer is used for detecting current in an electric circuit.
• The galvanometer is the device used for detecting the presence of small currents and voltage or for measuring their magnitude.
• The galvanometer is mainly used in the bridges and potentiometer where they indicate the null deflection or zero current.
• The potentiometer is based on the premise that the current sustaining coil is kept between the magnetic field experiences a torque.

EXPLANATION:

• From the above, it is clear that the galvanometer is the instrument used for detecting the presence of electric current in a circuit. Therefore option 1^st is correct.

​Additional Information 

Difference between Ammeter and Galvanometer:

• The ammeter shows only the magnitude of the current.
• The galvanometer shows both the direction and magnitude of the current.

The correct answer is Newton's first law.

CONCEPT:

• Newton’s first law of motion: It is also called the law of inertia. Inertia is the ability of a body by virtue of which it opposes a change.
• According to Newton’s first law of motion, an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force.
• The inertia of rest: When a body is in rest, it will remain at rest until we apply an external force to move it. This property is called inertia of rest.
• The inertia of motion: When a body is in a uniform motion, it will remain in motion until we apply an external force to stop it. This property is called inertia of motion.

EXPLANATION:

• When a bus suddenly starts moving, the passengers fall backward due to the law of inertia of rest or 1st law of Newton.
• Because the body was in the state of rest and when the bus suddenly starts moving the lower body tends to be in motion, but the upper body still remains in a state of rest due to which it feels a jerk and falls backward. Hence option 1 is correct.

Additional Information

Laws of Motion given by Newton are as follows:

Option 4 is correct

CONCEPT:

• Electric potential (V): The amount of work done to move a unit charge from a reference point (or infinity) to a specific point in an electric field without producing an acceleration is called electric potential at that point.

\({\rm{Electric\;potential\;}}\left( {\rm{V}} \right) = \frac{{{\rm{Work\;done\;}}\left( {\rm{W}} \right)}}{{{\rm{Charge\;}}\left( {\rm{q}} \right)}}\)

• Electrostatic Potential Energy: The amount of work done to move a charged particle from infinity to a point in an electric field is known as the potential energy of that charged particle.

CALCULATION:

Given that:

Electric charge (q) = 5 C

Potential difference (V) = 16 V

Work done (W) = charge (q) × potential difference (V)

Work done (W) = 5 × 16 = 80 J

The correct answer is 2 kJ.

CONCEPT:

• Potential energy: The energy of any object due to its position with respect to a reference point is called potential energy. It is denoted by PE.

Potential energy is given by:

PE = m g h.

Here, PE is the Potential Energy, m is the mass, g is the acceleration due to gravity, and h is the height at which the object is placed

CALCULATION:

Given that: 

Mass (m) = 10 Kg

Height (h) = 20 m

P.E. = 10 x 10 x 20

P.E.= 2000 J

 P.E. = 2 kJ

• Kinetic energy: The energy due to the motion of the object is called kinetic energy. 
  • Kinetic energy (KE) = 1/2 (mv²)
  • Where m is mass and v is velocity. 
• Since the object is stationary (at rest) so the velocity is zero. Hence the kinetic energy of the object will be zero.
• Only the potential energy of the object will be there at the height.

The correct answer is 490J

CONCEPT:

• Work-energy theorem: It states that the sum of work done by all the forces acting on a body is equal to the change in the kinetic energy of the body i.e.,

Work done by all the forces = Kf - Ki

\(W = \frac{1}{2}m{v^2} - \frac{1}{2}m{u^2} = {\bf{Δ }}K\)

Where v = final velocity, u = initial velocity and m = mass of the body

CALCULATION:

It is given that,

Mass (m) = 20 kg

Final Velocity (v) = 7 m/s and initial velocity (u) = 0 m/s

According to the work-energy theorem,

⇒  Work done = Change in K.E

⇒  W = Δ K.E

Since initial speed is zero so the initial KE will also be zero.

⇒  Work done (W) = Final K.E = 1/2 mv²

⇒  W = 1/2 × 20 × 7²

⇒  W = 10 × 49

⇒  W = 490J

CONCEPT:

Coulomb's law in Electrostatics –

• Coulomb's law state’s that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

Force (F) ∝ q₁ × q₂

\(F \propto \;\frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 10⁹ Nm²/C²

EXPLANATION:

Given – q₁ = 2 × 10^-7 C, q₂ = 3 × 10^-7 C and r = 30 cm = 30 × 10^-2 m

Force is equal to

\(F = \left( {9{\rm{\;}} \times {\rm{\;}}{{10}^9}} \right)\times \frac{{2 \times {{10}^{ - 7}} \times 3 \times {{10}^{ - 7}}}}{{{{\left( {30 \times {{10}^{ - 2}}} \right)}^2}}}\)

\( \Rightarrow F = \frac{{54 \times {{10}^{ - 5}}}}{{900 \times {{10}^{ - 4}}}} = 6 \times {10^{ - 3}}N\)

The correct answer is Kudankulam.

• Kudankulam Nuclear Power Plant is the largest nuclear power station in India by capacity.

Key Points

• Kudankulam Nuclear Power Plant is located 650 km south of Chennai, in the Tirunelveli district of Tamilnadu, India.
• The power plant will have a combined capacity of 6000 Mega Watt upon completion.
• The Atomic Energy Commission was established in 1948 by the efforts of Dr. Homi Jahangir Bhabha, the father of Atomic Energy Research in India.
• India's first atomic research reactor 'Apsara' started working in Trombay (near Mumbai) but India's first Nuclear Power reactor was established at Tarapur in 1969.
• Production of nuclear energy requires uranium, thorium, and heavy water, Uranium is found in Jharkhand, Rajasthan, and Meghalaya.

CONCEPT:

• Equation of motion: The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

V = u + at

V2 = u2 + 2 a S

\({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

Where, V = final velocity, u = initial velocity, s = distance traveled by the body under motion, a = acceleration of body under motion, and t = time taken by the body under motion.

EXPLANATION:

Given that:

Initial velocity (u) = 0

Distance (S) = 20 m

Time (t) = 4 sec

Use \({\text{S}} = {\text{ut}} + \frac{1}{2}{\text{a}}{{\text{t}}^2}\)

20 = 0 + \(\frac{1}{2} \times a \times 4^2\)

acceleration = a = 20/8 = 2.5 m/s²

CONCEPT:

• When moving through a magnetic field, the charged particle experiences a force.
• When the direction of the velocity of the charged particle is perpendicular to the magnetic field:
  • Magnetic force is always perpendicular to velocity and the field by the Right-Hand Rule.
  • And the particle starts to follow a curved path.
  • The particle continuously follows this curved path until it forms a complete circle.
  • This magnetic force works as the centripetal force.
• Centripetal force (F_C) = Magnetic force (F_B)

​⇒ qvB = mv²/R

​⇒ R = mv/qB

where q is the charge on the particle, v is the velocity of it, m is the mass of the particle, B is the magnetic field in space where it circles, and R is the radius of the circle in which it moves.

EXPLANATION:

Given that particle has charge e; mass = m; and moves with a velocity v in a magnetic field B. So

• Centripetal force (FC) = Magnetic force (FB)

​​⇒ qvB = mv2/R

\(\Rightarrow R=\frac{mv}{qB}\)

\(\Rightarrow r=\frac{mv}{Be}\)

So the correct answer is option 1.